Recent content by user1616

Thanx anyway. I prooved it.

Well using it I just prooved it for the first column. For the next columns, in the previews theorem I should know that for every x' != x there is a y' != y. (!= is different) Can anybody tell me whether this is true or not? Also if it is, does anybody have any proof?

Yes there should be no loops. Anyway I read a theorem that may help. It says: We have 2 trees T,T'. For every edge (x) that belongs to T there is an edge (Y) that belongs to T' so that: (T-{x})U{y} and (T'-{y})U{x} are also trees. I'm trying to use it...

well A spanning tree is a cohesive connected graph (T) with no loops that contains all the vertices of G... ...any idea?

Graph problem! Proove that: If we have a matrix |e11 e12 ... e1n| |e21 e22 ... e2n| | ..... | |en1 en2 ... enn| (eij edges of a graph G) where every row is a spanning tree of G then there is a recomposition of every row so that the columns are also spanning trees of G.